You have to prove that $e\in X\setminus\bar{A}$, if an only if there exists a neighbouhood $U$, for that $A\cap U= \varnothing $.

$\Rightarrow $ If $e\in X\setminus\bar{A}$, then it's clear that the nonempty set $X\setminus\bar{A}$ is open and contains $e$. So there exists a neighbourhood $U$ of $e$, s.t. $U\subset X\setminus \bar{A}$ and $A\cap U= \varnothing $.

$\Leftarrow $ Suppose that there is a neighbourhood $U$ of $e$ in $X$, s.t $A\cap U= \varnothing $, then we get $C= X\setminus U$ must be closed, $e\in X\setminus C$ and $A\subset C$. You know that $\bar{A}$ is the smallest closed set countaining $A$, so we have $\bar{A}\subset C$ and thus $e\in X\setminus \bar{A}$.